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Let $$x(t) = \frac{\sin t}{t} \qquad\text{and}\qquad y(t) = \frac{\sin 2t}{t}$$

  • Is it possible to find a LTI system such that $\mathcal{T}\{x(t)\} = y(t)$?
  • If not, what's the reason for that?

My try:

Assuming frequency response exists

$$Y(j\omega) = H(j\omega)X(j\omega) \tag{*}$$

Since $$\mathcal F\left\{\frac{\sin Wt}{\pi t}\right\} =\begin{cases} 1 &|\omega|\lt W \\0 &|\omega|\gt W\end{cases}$$ We have $$X(j\omega) =\begin{cases} \pi &|\omega|\lt 1 \\0 &|\omega|\gt 1\end{cases}$$ And $$Y(j\omega) =\begin{cases} \pi &|\omega|\lt 2 \\0 &|\omega|\gt 2\end{cases}$$ So it's not possible to find $H(j\omega)$ such that $(^*)$ holds. Because if $|\omega| \gt 1$ we have $X(j\omega) = 0$ which implies $Y(j\omega) = 0$ contradicting $Y(j\omega) = \pi$ for $1\lt |\omega| \lt 2$.

  • Is my argument right?
  • Also how we can prove the impossibility for the case frequency response doesn't exist? What's the intuition for this? I mean is it intuitively clear that we can't have $\mathcal{T}\{x(t)\} = y(t)$ for LTI system?
Gilles
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S.H.W
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  • You are correct. For an argument as to why in the opposite direction (that is, $\mathcal T{x(t)]=x\left(\frac t2\right)$ instead of $\mathcal T{x(t)]=2x(2t)$ as in your question), a linear but not time-invariant system will work, see this recent question and its answers – Dilip Sarwate Jun 04 '20 at 21:00
  • @DilipSarwate Thanks for your reply. How we can prove that when frequency response doesn't exist? – S.H.W Jun 04 '20 at 21:52
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    Do you have an example of an LTI system for which the frequency response does not exist in mind? – Dilip Sarwate Jun 05 '20 at 13:56
  • @DilipSarwate Yes, take a look: https://dsp.stackexchange.com/questions/67302/determining-output-of-a-lti-system I think we can construct other non-stable systems as well such that frequency response doesn't exist. – S.H.W Jun 05 '20 at 16:55
  • No such LTI system can do this for you. You have a divide-by-zero problem. for $ 1 < \omega < 2$, your LTI system has to create non-zero energy at frequencies of where zero energy is input. – robert bristow-johnson Jun 05 '20 at 17:54
  • @robertbristow-johnson That's right. I wonder what would happen if frequency response doesn't exist. How we can prove that $\mathcal{T}{\frac{\sin t}{t}} = \frac{\sin 2t}{t}$ is impossible for any LTI system using only basic properties without resorting to Fourier transform to avoid the possibility that frequency response diverges? – S.H.W Jun 07 '20 at 20:53
  • you prove it by contradiction. division-by-zero leads to contradiction. i.e. dividing a non-zero number by zero leads to contradiction. – robert bristow-johnson Jun 08 '20 at 16:15
  • @robertbristow-johnson That works in the case frequency response supposed to exist. I mean proving using only linearity and time-invariability. – S.H.W Jun 08 '20 at 16:55
  • do you know how to prove, given *only* L and TI, how to prove that the output $y(t)$ is the convolution of the input $x(t)$ and some "$h(t)$" that we can interpret as an impulse response? can you do that? then try this for an input: $$ x(t) = A \cos(\omega t + \theta) $$ and see what comes out. – robert bristow-johnson Jun 08 '20 at 17:20

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